If (0, 0) and (2, 0) are the two vertices of a triangle whose centroid is (1, 1), then the area of the triangle is: -Maths 9th

If (0, 0) and (2, 0) are the two vertices of a triangle whose centroid is (1, 1), then the area of the triangle is: -Maths 9th
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(b) \(\bigg(rac{2\sqrt{13}+20\sqrt2}{\sqrt{13}+\sqrt{17}+5\sqrt2},rac{8\sqrt{13}-6\sqrt{17}}{\sqrt{13}+\sqrt{17}+5\sqrt2}\bigg)\)Let A(x1, y1), B(x2, y2), C(x3, y3) be the vertices of ΔABC the coordinates who mid-points are given as P(1, 1), Q(3, 4) and R(2, –3)Now, \(rac{x_1+x_2}{2}=1\) and \(rac{y_1+y_2}{2}=1\)              ....(i) \(rac{x_2+x_3}{2}=3\) and \(rac{y_2+y_3}{2}=4\)              ....(ii) \(rac{x_1+x_3}{2}=2\) and \(rac{y_1+y_3}{2}=3\)              ....(iii)∴ \(rac{x_1+x_2+x_2+x_3+x_1+x_3}{2}=6\) and \(rac{y_1+y_2+y_2+y_3+y_1+y_3}{2}=2\)⇒ x1 + x2 + x3 = 6 and y1 + y2 + y3 = 2Now, (x1 + x2 + x3) – (x1 + x2) = 6 – 2 ⇒ x3 = 4 and (y1 + y2 + y3 ) – (y1 + y2) = 2 – 2 ⇒ y3 = 0 (x1 + x2 + x3) – (x2 + x3) = 6 – 6 ⇒ x1 = 0 and (y1 + y2 + y3) – (y2 + y3) = 2 – 8 ⇒ y1 = – 6 (x1 + x2 + x3) – (x1 + x3) = 6 – 4 ⇒ x2 = 2 and (y1 + y2 + y3) – (y1 + y3) = 2 + 6 ⇒ y2 = 8 ∴ A(0, – 6), B(2, 8), C(4, 0) are the vertices of ΔABCNow, a = BC = \(\sqrt{(4-2)^2+(0-8)^2}\) = \(\sqrt{4+64}\) = \(\sqrt{68}=2\sqrt{17}\)b = AC = \(\sqrt{(4-0)^2+(0+6)^2}\) = \(\sqrt{16+36}\) = \(\sqrt{52}=2\sqrt{13}\)c = AB = \(\sqrt{(2-0)^2+(8+6)^2}\) = \(\sqrt{4+196}\) = \(\sqrt{200}=10\sqrt{2}\)∴ Co-ordinates of incentre of Δ ABC are
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Description : ABCD is a trapezium where AB and CD are non-parallel sides. If the vertices A, B, C and D are concyclic, then -Maths 9th

Last Answer : answer:

If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Description : If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Last Answer : (c) RhombusCo-ordinates of P are \(\bigg(rac{-1-1}{2},rac{-1+4}{2}\bigg)\)i.e, \(\big(-1,rac{3}{2}\big)\)Co-ordinates of Q are \(\bigg(rac{-1+5}{2},rac{4+4}{2}\bigg)\)i.e, (2, 4)Co-ordinates of R ... \sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

(0, –1) and (0, 3) are the two opposite vertices of a square. The other two vertices are: -Maths 9th

Description : (0, –1) and (0, 3) are the two opposite vertices of a square. The other two vertices are: -Maths 9th

Last Answer : (c) (15, 19)Let Δ(a, b) be the fourth vertex of the parallelogram ABCD. The diagonals of a parallelogram bisect each other at point O (say), so the diagonals AC and BD have the same mid-point.

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : NEED ANSWER